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mamaluj [8]
3 years ago
7

an object which has a mass of 70 kg is sitting on a cliff 10 m high calculate the objects gravitational potential energy given g

= 1o/s^
Physics
1 answer:
Alex73 [517]3 years ago
5 0
GPE = mgh (for objects close to earth where g can be considered to be constant) Where m is the mass, g is the acceleration due to gravity (which the question has told us to assume is 10ms^-2), and h is the height of the object.

Substituting the values in:
GPE = 70*10*10 = 7000J
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The critical angle for a special type of glass in air is 30.8 ◦ . the index of refraction for water is 1.33. what is the critica
Alina [70]
When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted light, and all the light is reflected. The value of this angle is given by
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n2 and n1 are the refractive indices of the second and first medium, respectively.

In the first part of the problem, light moves from glass to air (n_a=1.00) and the critical angle is \theta_c = 30.8^{\circ}. This means that we can find the refractive index of glass by re-arranging the previous formula:
n_g=n_1 =  \frac{n_2}{\sin \theta_c}= \frac{1.00}{\sin 30.8^{\circ}}=1.95

Now the glass is put into water, whose refractive index is n_w = 1.33. If light moves from glass to water, the new critical angle will be
\theta_c = \arcsin ( \frac{n_2}{n_1} )=\arcsin( \frac{n_w}{n_g} )=\arcsin( \frac{1.33}{1.95} )=\arcsin(0.68)=43.0^{\circ}
6 0
3 years ago
A 1700 W laser emits light with a wavelength of 700 nm into a 3.0 mm diameter beam. What force does the laser beam exert on a co
maks197457 [2]

Answer:

F=5.7×10⁻⁶

Explanation:

Not knowing a formula outright, I decided to follow the units of some relationships I did know. Radiation pressure is defined as force per area and also intensity divided by velocity (the speed of light here of course). Breaking intensity down into power per area and isolating force gave me the relationship F=(Power/Velocity),where power is given and the velocity is a constant.

My work is in the attachment, where I double checked the units too, comment with any questions.

5 0
3 years ago
A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov
Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

4 0
3 years ago
Read 2 more answers
During an auto accident, the vehicle’s air bags deploy and slow down the passengers more gently than if they had hit the windshi
Allushta [10]

Answer:

d = 0.38 m

Explanation:

As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:

vf = v₀ -a*t  

If vf = 0, we can solve for v₀:

v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s

With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.

Just for simplicity, we can use the following equation:

vf^{2} -vo^{2} = 2*a*d

where vf=0, v₀ =21.2 m/s and a= -588 m/s².

Solving for  d:

d = \frac{-vo^{2}}{2*a} = \frac{(21.2m/s)^{2} }{2*588 m/s2} =0.38 m

⇒ d = 0.38 m

5 0
3 years ago
Read 2 more answers
In a given reversible process, the temperature of an ideal gas is kept constant as the gas is compressed to a smaller volume. Wh
sesenic [268]

Answer:

B. The gas must release heat to its surroundings.

Explanation:

Since the process is reversible , it must be slow . Since temperature is kept constant so the internal energy is constant , Since volume is decreased , pressure must be increased. Since work is done on the gas ( gas is compressed )  , so heat must be released by the gas so that its internal energy remains constant.

7 0
3 years ago
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